Optimizing the Controller Design to Guide the Motion of a Maglev Train

Transcription

Optimizing the Controller Design to Guide
the Motion of a Maglev Train
The challenge: To design a robust feedback controller and optimize the control
parameters to guide the motion of a Maglev train along the guideway
© Maplesoft, a division of Waterloo Maple Inc., 2008
Introduction
Problem Definition
1. Model Development
1.1 Linear Model of the Magnet
1.2 Model for the Linearized Magnet Force
1.3 Model of Magnet Dynamics
2. Control System Design
2.1 PI Controller
2.2 Control Specifications and
Measurements
2.3 Control System Transfer Function
2.4 Controller Gains & Closed Loop System
2.5 The Final Design
3. Vehicle Model
3.1 Equations of Motion
3.2 Response Plots
Results
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Introduction
Magnetically Levitated (Maglev) trains differ from conventional trains in that they are levitated, guided and
propelled along a guideway by a changing magnetic field rather than by steam, diesel or electric engine. The
absence of direct contact between the train and the rail allows the Maglev to reach record ground
transportation speeds, which are on par to that of commercial airplanes.
Figure 1: (A) Cross sectional view of cabin & (B) Expanded view of magnet support structure
An Electromagnetic Suspension (EMS) Maglev train uses
the attractive forces of magnets (& electro-magnets),
positioned below a guideway, to levitate a train and guide
it along the guideway (fig 1).
Three phase AC propulsion coils mounted along the steel
rails of the guideway provide a moving magnetic field that
interacts with the periodic magnetic field created by the
lifting magnets to propel the train forward. There are 24
lifting magnet pairs (48 magnets all together) mounted
along the length of the train. The lifting magnets, which
are canted at an 37° angle towards the steel rails,
consists of high temperature superconducting (HTS)
magnets and conventional (non-superconducting) coils
(fig 2). A periodic magnetic field is generate by aligning
the lifting magnet pairs so that their north and south poles
alternate.
Figure 2: Expanded view of lifting magnets
There are many advantages and disadvantages
associated with Maglev trains in comparison to
conventional steel wheel on steel rail trains . The table
(table 1) below highlights some of these.
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Advantages
Disadvantages
Record breaking ground transportation speeds
reaching 500 km/h (300 mph)
High infrastructure costs
Less expensive to operate and maintain than
traditional high-speed trains or even planes, and
as a consequence lower cost per passenger per
mile
Not compatible with conventional wheeled tracks
making it limited to only those places where
maglev lines run.
Silent at low speeds
Increased noise pollution at high speeds
Significant energy savings because the vehicle
does not have to carry around the weight of its
propulsion system
No associated maintenance cost because there
are no parts that are in contact with ground
Improved ride comfort
Table 1: Advantages and Disadvantages of Maglev trains over conventional trains
Problem Definition
EMS Maglev trains require robust controllers to guide the train along the guideway. The controller prevents the
train from colliding into the guideway by maintaining a constant air gap between the train and guideway
regardless of inconsistencies in the guideway, change in direction or angle of the track, and environmental
forces such as wind.
Control of the vehicle is achieved by a combination of passive (i.e. through the orientation of the magnets
along the vehicle) and active (i.e. with an active control system) techniques. Each magnetic coil has its own
independent control system. This independence allows the pitching (forward and aft rotations) to be controlled
by the magnets at the front and rear of the vehicle acting independently to maintain the air gap at 2 inches.
Because the magnets are canted at 37°, the lateral and yaw motions (motions that turn the vehicle left or right
relative to the direction of travel) are also controlled automatically when the air gaps are maintained at the
nominal 2 inches. For example, if a wind gust were to shift the vehicle to the right, the gap on the left side
would become smaller and the one on the right side would become bigger. Each of the magnets would then
correct for the motion by returning the magnets back to equilibrium.
Lift (up and down motion) is controlled when the left magnet and right magnet move in unison. For example,
when the rail rises it causes the gap on both the right and left side of the vehicle to get smaller. The magnet
control systems respond to this change by adjusting the current flowing through the magnets to return the gap
to the desired 2 inch setting.
The control for roll (rotations to the left and right along the direction of travel) is done passively by offsetting
each of the magnets pairs by 1 inch. This offset acts to provide a magnetic spring that naturally restores the
vehicle so it is perpendicular to the guideway whenever a disturbance occurs.
This worksheet will examine the design of a robust feedback controller to guide the motion of a EMS Maglev
train along the guideway.
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1. Model Development
1.1 Creating a Linear Model of the Magnet
The force from the magnet is a function of the gap distance z and the current in the coils Ic. The force can
be developed from first principles using Maxwell's equations, or by using the inductance of the magnets,
but because of the iron in the core, the actual force is a complex function of the geometry of the magnet,
the amount of iron in the magnet and the rail, and the inherent hysteresis of the iron. The nonlinear effects
of the magnet can be approximated to first order so that the magnet force is proportional to the square of
the current flowing in the magnet coils and inversely proportional to the square of the gap between the rail
and the magnet. Thus:
f d z, Ic /K
1 2
Ic
z2
z, Ic /
K Ic2
(1)
2
z
If we assume that the vehicle mass is m and the gravitational acceleration is g, then a simple free body
model of the vertical motion gives the differential equations for the motion as follows.
Velocity:
ode1 d
d
f z t , Ic t
Fd t
v t Cg =
C
dt
m
m
d
K Ic t 2
Fd t
v t Cg =
C
2
dt
m
z t m
(2)
Displacement:
ode2 d
d
z t =v t
dt
d
z t =v t
dt
(3)
Suppose the weight of each car in a two car Maglev train is 135,000 pounds. Using the same units, the
acceleration of gravity is defined as 32.2 ft/s/s. For now let's assume the proportionality constant K is unity.
This then gives the parameters for the model as:
params d K = 1, g = 32.2, m =
135000
32.2
:
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1.2 Creating a Model for the Linearized Magnet Force
The first step in obtaining an accurate model of the vehicle is to model the nonlinear magnet dynamics.
Therefore, we will create, using the DynamicSystems package, the nonlinear differential equation model,
using the coupled differential equations ode1 and ode2 defined above, with the parameters defined above.
ode1 d eval ode1, params
d
0.0002385185185 Ic t
v t C32.2 =
dt
z t 2
2
C0.0002385185185 Fd t
(4)
d
z t =v t
dt
sys_de d DynamicSystems DiffEquation
outputvariable = z t , v t
:
(5)
ode1, ode2 , inputvariable = Ic t , Fd t ,
To see what the object sys_de looks like we use the PrintSystem command in the
DynamicSystems package.
DynamicSystems PrintSystem sys_de
Diff. Equation
continuous
2 output(s); 2 input(s)
inputvariable = Ic t , Fd t
.
0.0002385185185 Ic t
de = v t C32.2 =
z t 2
2
(6)
.
C0.0002385185185 Fd t , z t
=v t
We now have a differential equation system object with the inputs Ic and Fd, and outputs z and v.
The proportionality constant K in equation (1) is derived from the fact that the magnetic force is given by
the cross product of the B field with the current producing the magnetic field. Since the current produces
both the lifting field and the field it interacts with (through the closed loop around the C shaped magnet and
the rail), the lifting force is proportional to the square of the B field over the area A of the magnet's pole
face. The proportionality constant for this force is
1
. Thus the force from the magnet is:
2µ0
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F=
B2A
:
2µ0
The B field is given by µ$H where µ is the magnetic permeability of the iron in the magnet and rail (due to
the hysteresis of the iron, this term is not constant) and the magnetic flux is given by the current in the coil
multiplied by the number of turns in the coil. Thus, the constant K is given by:
2 2
K=
µN A
:
2µ0
1.3 Using the Model to Develop an Understanding of the
Magnet Dynamics
Based on a simple understanding of the physics, it should be clear that when there is no current in the
coils the vehicle will drop under the influence of gravity causing the undercarriage to hit the guideway.
Similarly, when the current in the magnet is very large, the magnet will be attracted to the rail and hit the
rail (in fact the assumed force function is infinite when the gap z is zero). Therefore, as we analyze the
nonlinear differential equation we should expect that the dynamics are unstable.
To investigate the dynamics, the force term f z, t must be linearized. The basic superconducting magnet
was designed to have a nominal value of N$I equal to 50,000 ampere turns. The control coils, which
regulate the current through the superconducting magnets, were designed to allow a maximum of 15,000
ampere turns. In addition, the nominal air gap for the system is 2 inches. These values establish the
numerical value for force and therefore for the term K. The linearization of the nonlinear term in (1), from
the Taylor series expansion around the nominal 2 inch gap and 50,000 ampere turns current in the coils, is
obtained from the partials Kz =
v
v
f z, Ic : and KI =
f z, Ic :
vz
v Ic
2
v
2 K Ic
f z, Ic = K 3
vz
z
and
v
2 K Ic
f z, Ic =
v Ic
z2
Ic and z are substituted into the expansion, the linearization becomes (in this
and the subsequent analysis, the variations in z and Ic, ∆z and ∆Ic, are written as z and Ic respectively):
When the nominal values of
f z t , Ic t
= Kz$z t CKI$Ic t :
The nominal value for the vehicle mass, and the nominal value of K gives Kz as 281,000 pounds/foot and
the nominal value of KI as 3372000 pounds/kiloampere-turn.
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This then provides the linear dynamics as:
ode3 d
d
Kz$z t
KI$Ic t
Fd t
v t =
C
C
dt
m
m
m
:
params d K = 1, m = 50500 , g = 32.2, Kz = 281000, KI = 3372000 :
Now we evaluate the equation at the operating point derived above:
562
6744
1
d
v t =
z t C
Ic t C
Fd t
101
101
50500
dt
(7)
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2. Control System Design
The analyses in the previous section allow us to begin the design of the control system. The first step in this is
an investigation of the underlying dynamics in order to understand how best to control the vehicle. In this
section we will use the equations defined above to develop a controller which will effectively control the
behavior of our system to external perturbations, such as changes in guideway and disturbance forces (such
as wind)), according to a set of well defined control criteria.
To start, we define the differential equation system object that we will be working with using ode2 and ode3
above:
sys_de d DynamicSystems DiffEquation
:
ode2, ode3 , inputvariable = Ic t , Fd t ,
Diff. Equation
continuous
inputvariable = Ic t , Fd t
(8)
.
.
562 z t
6744 Ic t
Fd t
de = z t = v t , v t =
C
C
101
101
50500
For the design of the control system, the transfer function poles and zeros for the linear differential equation
need to be derived. We start by using DynamicSystems to get the ZeroPoleGain version of sys_de:
sys d DynamicSystems ZeroPoleGain sys_de
Zero Pole Gain
continuous
(9)
inputvariable = Ic s , Fd s
outputvariable = z s , v s
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Zero-Pole-Gain Analysis
Zero-Pole Plot
Zeros
sys:-z =
Zero-Pole Plot
1
0
0
0.5
Poles
K2
sys:-p =
1
1
K
56762 ,
56762 ,
101
101
1
1
K
56762 ,
56762 ,
101
101
1
1
K
56762 ,
56762 ,
101
101
1
1
K
56762 ,
56762
101
101
K1
0
1
2
K0.5
K1
Figure 3: Zero-Pole Plot
Gains
sys:-k =
6744
101
1
50500
6744
101
1
50500
From the zero-pole plot we see immediately that there are two poles: one in the right half plane and one in the
left. As we deduced from first principles, the system is unstable. Just for completeness, if only the the transfer
function for the system were needed, it would be obtained from:
sys d DynamicSystems TransferFunction sys_de
Transfer Function
continuous
(10)
inputvariable = Ic s , Fd s
outputvariable = z s , v s
outd sys:-tf
6744
101 s2 K562
1
50500 s K281000
6744 s
101 s2 K562
s
2
50500 s K281000
2
(11)
9 of 25
Now we can design the control system.
2.1 Using the Superconductor for Integral Control for a PI
Controller
Superconductor magnets do not like rapidly varying currents. We can use the superconductor as an
integrator (this comes from physics). The magnet has essentially zero resistance, so the differential
equation for the magnet is:
Lsc
d
Icsc t = v t :
dt
Thus, for any voltage v t applied to the superconductor, the current in the magnet is the integral of the
voltage. Making the control signal the voltage, the superconductor will integrate this control and the gain on
this signal will be the integral control gain, which we denote by Ki.
Similarly, the control coil has the model:
Lc
d
Ic t CRcIc t = v t :
dt
and of course in steady state the current in the control coil is simply
v t
multiplied by the proportional
Rv
control gain (Kp).
This analysis shows that the control system is a proportional plus integral controller with gains Kp and Ki,
but we still have not addressed how to stabilize the lift magnets. Let's investigate the measurements that
will be required to stabilize the magnets.
2.2 The Control Specifications and the Measurements Needed
When a Maglev "flies" above the guideway (and below the rails in this design), it is moving along in inertial
coordinates that are only slightly constrained by the proximity of the guideway. This means that a slight
change in the rail location relative to the vehicle - one that is caused by an irregularity in the rail for
instance - can be ignored. Thus, measurements of the gap between the magnet and the rail are not
sufficient to tell the control system what to do. In most attractive Maglev designs, therefore, a
measurement of the gap is supplemented with a measurement of the acceleration of the magnet in the
direction of the rail (remember in this design the magnets are canted at an angle so the force is not applied
purely along the vertical (z) axis). The control system is therefore required to null accelerations.
Acceleration alone can not be the only feedback. The guideway (and the rail along with it) must follow the
terrain and the desired route (turning left and right). This means that the gap between the magnets and the
rail must be measured too. Furthermore, the availability of acceleration as a measurement would allow
derivative control (i.e. the control could be full state feedback PID control) simply by integrating the
acceleration.
The control criteria is therefore:
1. At a fast scale (response times between 0.1 s and 1 s), zero the acceleration.
10 of 25
The motivation for this is twofold. First, this cancellation will result in closed loop dynamics that are second
order, and secondly the bandwidth of the closed loop servo can be set to any desired value (independent
of the magnet dynamics). As a second order system, the gains for the measurements and the proportional
feedback can be selected to make the magnet response optimal in the LQ sense (i.e. we can select the
closed loop bandwidth to be 10 Hz as required in the specifications, and we can set the damping to the
optimal value of
requires that
Ki
2
2
2
) . The cancellation of the zero with the left half plane pole of the unstable dynamics
=
Kz
:
m
The following simplifications are done to get the closed loop transfer function into the required form:
Kz
, Gcl s
m
factor algsubs Ki = sqrt
Kz
m
sC
Kf KIc m Ka s3 CKf KIc m Ka s2
Kf KIc m
Kz
m
Kz
CKf KIc m s CKf KIc m
m
2
CKf KIc m Kv s CKf KIc m Kv s
(14)
Kz
3
Cs m
m
Ks Kz
Note that the possibility of having a common factor in the numerator and denominator is not obvious, but it
is possible to select the values of Ka, Kf and Ki to make the denominator of this transfer function equal to
sC
Kz
m
2
s2 C 2ζCL$ωCL$s C ωCL : , where the values of the desired closed loop denominator
are:
ζCL =
2
2
ωCL = 20$π :
:
To simplify the algebra, let's assume that the zeros introduced by the acceleration gain and velocity gain
are both at the same location denoted by ωz. This assumption makes the acceleration and velocity gains
as follows:
Ka =
1
2
ωz
:
and
Kv =
1
:
ωz
While it might seem like a large set of simultaneous nonlinear equations need to be solved to completely
specify the closed loop dynamics, we have systematically been eliminating the non-linearities in the
solution as we constrained the gains. Therefore, the values for the gains are completely specified by the
location of the zeros, which because of our assumption is ωz. Define the open loop undamped natural
frequency as ω
OL
=
Kz
:
m
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ωCL 1 C
1C
Then ω
=
z
ωOL
ωCL
2$ζCL C
$ 2$ζCL C
ωOL
ωCL
ωOL
:
ωCL
The values for the feedback gains Ka and Kv and the integral gain Ki are completely determined from the
above, and the value of Kf is:
ωCL
Kf =
KIc$ 1 C
ωCL
2
:
ωz
2.5 The Final Design
We can now verify that the design is correct by examining the system response to disturbance forces and
guideway changes and errors. Toward that end, let's find the numerical values for the gains based on the
equations we have derived above.
Substituting ζCL, ωCL, and ωOL from the specification completely determines the gains that, along with the
other parameters, can be used to create the system model that we want to examine.
Thus the parameters are:
params d m = 4192.5 , Kz = 3372000 , KIc = 17984, Ka = 0.00015866 , Kv = 0.0252, Ki
= 28.3599, Kf = 2463.2 :
The state space model for the complete linear system can be written from the block diagram in Figure 3, in
the following manner:
We start at the right side of the diagram, moving from integrator to integrator for each of the states. The
first integrator (for the velocity) gives:
ode1 d
d
x t =v t :
dt
The next integration (of the acceleration) is a little complex because there is an algebraic loop involved.
Thus the differential equation for the acceleration has a term on the right that involves the derivative of the
velocity (i.e. the acceleration) that must be removed by bringing the term to the right and dividing through
by the resulting coefficient (which is 1 C
d
ode2 d
v t =
dt
Kf$Ka$KIc
), so:
m
KzKKf$KIc
Kf$Kv$KIc
m
m
$x t K
$v t
Kf$Ka$KIc
Kf$Ka$KIc
1C
1C
m
m
14 of 25
Kf$KIc
Kf$KIc
m
m
C
$x3 t C
$g t
Kf$Ka$KIc
Kf$Ka$KIc
1C
1C
m
m
Fd t
C
Kf$Ka$KIc
m 1C
m
d
Kz KKf KIc x t
Kf Kv KIc v t
Kf KIc x3 t
v t =
K
C
dt
Kf Ka KIc
Kf Ka KIc
Kf Ka KIc
m 1C
m 1C
m 1C
m
m
m
Kf KIc g t
Fd t
C
C
Kf Ka KIc
Kf Ka KIc
m 1C
m 1C
m
m
(15)
The last equation comes from the integrator in the integral compensation, and this also is complex
because the right hand side also has the acceleration in it. We let Maple do the work of solving for this and
eliminating it as follows:
ode3 d
d
d
x3 t = KKi$Ka$
v t KKv$Ki$v t KKi$x t CKi$g t :
dt
dt
We now substitute all of the parameter values into these equations:
d
x t =v t
dt
(16)
d
v t = K3647.333890 x t K99.48571587 v t C3947.845868 x3 t
dt
C3947.845868 g t C0.0002385211688 Fd t
(17)
d
x3 t = K0.004499581734
dt
C28.3599 g t
(18)
d
v t
dt
sys_de d DynamicSystems StateSpace
outputvariable = x t , v t
:
K0.71466948 v t K28.3599 x t
ode1, ode2, ode3 , inputvariable = Fd t , g t ,
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State Space
continuous
2 output(s); 2 input(s); 3 state(s)
inputvariable = Fd t , g t
outputvariable = x t , v t
statevariable = x1 t , x2 t , x3 t
a = K99.48571587, K3647.333890, 3947.845868 ,
1, 0, 0 ,
K0.2670253700774340814, K11.94842305075683474, K17.76365515630017511
(19)
0.0002385211688
b=
3947.845868
0
0
K1.073245494304810699 10
c=
d=
K6
10.59624484369982489
0 1 0
1 0 0
0 0
0 0
We can verify the behavior of our controller by examining how it responds to disturbances and guideway
changes and/or irregularities. If we want to see what happens when there is a change in the force applied
to the vehicle (as for, example, a vertical gust), the Fd(t) input can be set to a step input, and if we want to
see what happens when there is an downward slope, the guideway input g t can be set to be a ramp.
16 of 25
Control Response to Disturbance Forces and Guideway Changes
Slope
Slope [inches]
0
2
4
6
8
time [sec]
K3
K5
K7
10
Gap Error [inches]
The response obtained below was calculated for a
Generate Response
10,000
Plotpound aerodynamic gust of wind, and for a
guideway slope that descends 1 ft over 10 seconds
of travel.
Gap Error
0.025
0.020
0.015
0.010
0.005
0
K9
Figure 6A: Input Signal
0
2
4
6
8
time [sec]
10
Figure 6B:
System Response Plot
A quick look at the control response plots shows that the controller is able to correct for air gap changes
due to disturbance forces, such as a 10,000 pound aerodynamic gust of wind, in about 0.1 seconds.
Moreover, the air gap due to such a strong wind is only 0.004 inches. In addition, we can see that the
controller is able to maintain the air gap even when the guiderail path is perturbed. These results show that
the controller meets the design criteria specified in Section 2.2.
3. Vehicle Model
By maintaining the air gap between the guideway and train we expect the control system to be able control the
motion of the vehicle in terms of heave, sway, pitch and yaw. In this section we will use the magnet model and
control system equations determined previously to derive the equations of motion for the entire Maglev
system. We will then examine the Maglev train system response to disturbance forces and guideway changes.
3.1 Equations of Motion
Figure 7 shows the magnets as they are grouped along the sides of the vehicle. Twelve groups of four
magnets are mounted along the vehicle (4 magnets are grouped as a single assembly and the force they
apply is along the axis of the magnets through a compliant joint). Because of the cant angle the vertical
forces (along the z-axis) are given by f i cos
β and the horizontal forces along the y axis are given by
17 of 25
f i sin β , where β is the magnet cant angle.
Figure 7. Maglev train geometry
Location of the 12 magnet modules, their produced forces f1 through f12, and the three lever
arms for the moments (magnets are placed symmetrically so these 3 define all 6 lever arms).
The moments and forces created by the magnets depend, of course, on the geometric location of the
magnets. All of the magnets are canted at an angle of 35°. Thus the attractive force of the magnet (which
is what has been modeled so far) must be resolved into vertical and sideways (sway) forces. Once this has
been done, the moments created by the magnets can be defined. We create a vector of forces for the
magnets and then also create a vector of lever arms for each of the rotational degrees of freedom. When
defining the equations we assume that all of the rotational angles are small so the gyroscopic forces and
Coriolis forces can be neglected. The equations of motion then become:
18 of 25
Equations of Motion:
Iθ $
d
dt
d
θ t
dt
= LΘf$$ cos β
2
Iθ
m
= LΘ f cos β
Θ
d
ψ t
2
dt
Iψ
(20)
d
d
y t = Lyf$$ cos β
dt
dt
2
d
y t = Ly f cos β
(22)
2
dt
The row vectors L
d
ψ t
dt
= Lψf$$ sin β
2
d
θ t
2
dt
m$
d
dt
Iψ$
m$
m
= Lψ f sin β
(21)
d
d
z t = Lzf$$ sin β
dt
dt
2
d
z t = Lz f sin β
(23)
2
dt
, Lψ, Ly, and Lz are each an array of distances (or +1's and -1's in the case of the
forces) that multiply the forces to give the forces or moments (summed together at the center of gravity)
for the 12 magnet modules. They are given by:
Row Vectors L , L , L , and L :
Θ
ψ
y
z
1 .. 12 Vectorrow
LΘ =
Ly =
1 .. 12 Vectorrow
Data Type: anything
:
Storage: rectangular
Data Type: anything
Lψ =
Order : Fortran_order
1 .. 12 Vectorrow
1 .. 12 Vectorrow
Data Type: anything
:
Storage: empty
Lz =
:
Data Type: anything
:
The equations of motion defined above are combined with the magnet module equations to obtain the
closed-loop system equations that describe the behavior of the Maglev train.
We rename the state variable x3(t) to aux(t):
contorlODEs d subs x3 t = aux t , ode1, ode2, ode3
d
d
x t =v t ,
v t = K3647.333890 x t K99.48571587 v t
dt
dt
C3947.845868 aux t C3947.845868 g t C0.0002385211688 Fd t ,
(24)
d
aux t
dt
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= K0.004499581734
d
v t
dt
K0.71466948 v t K28.3599 x t
C28.3599 g t
Next, the equations are duplicated 12 times for each of the magnet modules:
listVarnames d x, v, aux, g :
allcoils d seq op subs map dummy/dummy = dummy i, listVarnames , contorlODEs , i
= 1 ..12 :
Expression for the force generated per module (for n modules) is defined as:
coilForceExpr d F t =
m$ diff v t , t
n
m
F t =
d
v t
dt
n
(25)
Parameter values are substituted:
coilForce d subs
n = 12 , params, coilForceExpr
d
F t = 349.3750000
v t
dt
(26)
Creating the force expressions for y and z axes:
YForceExpr d Fy t = F t $sin β
Fy t = F t sin β
(27)
ZForceExpr d Fz t = F t $cos β
Fz t = F t cos β
(28)
Substituting parameter values and generating the force for each individual module:
listVarnames d F, Fy, Fz, v :
YForce d map eval, seq subs map dummy/dummy = dummy i, listVarnames ,
subs coilForce, β = evalf convert 35,'units','deg','rad' , YForceExpr , i = 1 ..12
ZForce d map eval, seq subs map dummy/dummy = dummy i, listVarnames ,
subs coilForce, β = evalf convert 35,'units','deg','rad' , ZForceExpr , i = 1 ..12
:
:
20 of 25
Vehicle Dynamics
Heave
The vehicle dynamics in the z-direction is then given by summing up all of the Z forces:
odeZ d eval subs params, ZForce, m$diff z t , t$2 = add x, x = seq Fz
4192.5
d2
2
dt
z t
C286.1912455
C286.1912455
C286.1912455
C286.1912455
C286.1912455
= 286.1912455
d
dt
d
dt
d
dt
d
dt
d
dt
d
v1 t
dt
C286.1912455
i t , i = 1 ..12
d
v2 t
dt
(29)
d
v4 t
dt
d
v5 t C286.1912455
v6 t
dt
d
v7 t C286.1912455
v8 t
dt
d
v9 t C286.1912455
v10 t
dt
d
v11 t C286.1912455
v12 t
dt
v3 t
C286.1912455
Sway
The vehicle dynamics in the y-direction is then given by summing up all of the Y forces:
odeY d eval subs params, YForce, m$diff y t , t$2 = add x, x = seq
= 1 ..12
2
4192.5
d
y t
dt2
K200.3932674
K200.3932674
K200.3932674
K200.3932674
K200.3932674
= K200.3932674
d
dt
d
dt
d
dt
d
dt
d
dt
d
v1 t
dt
C200.3932674
K1 i$ Fy
d
v2 t
dt
i t ,i
(30)
d
v4 t
dt
d
v5 t C200.3932674
v6 t
dt
d
v7 t C200.3932674
v8 t
dt
d
v9 t C200.3932674
v10 t
dt
d
v11 t C200.3932674
v12 t
dt
v3 t
C200.3932674
21 of 25
Pitch
The dynamic equation is a function of the level arm distance from the center of gravity for each of the
modules.
Collecting the level arm into a list:
LPitch d K56.7, K56.7, K11.52, K11.52, K5.76, K5.76 :
LPitch d op LPitch , op map x/Kx, LPitch
:
The equation of motion of the pitching motion, with augmented forces generated by the motion of the
modules relative to the rail, is given by:
odeTheta d evalf subs
Iθ = 882500, β = convert 37,'units','deg','rad' , params, ZForce, Iθ
$diff θ t , t$2 = add x, x = zip `*`, LPitch, seq Fz
Ccot β $ψ t $add x, x = map x/x$x, LPitch
8.82500 105
d2
2
dt
= K16227.04362
θ t
K3296.923148
K1648.461574
C16227.04362
C3296.923148
C1648.461574
d
dt
d
dt
d
dt
d
dt
d
dt
d
v1 t
dt
i t , i = 1 ..12
K16227.04362
d
v4 t
dt
d
v5 t K1648.461574
v6 t
dt
d
v7 t C16227.04362
v8 t
dt
d
v9 t C3296.923148
v10 t
dt
d
v11 t C1648.461574
v12 t
dt
v3 t
Kθ t
d
v2 t
dt
(31)
K3296.923148
K13523.1120 θ t
K17945.77574 ψ t
22 of 25
Yaw
LYaw d 56.7, K56.7, 11.52, K11.52, 5.76, K5.76 :
LYaw d op LYaw , seq op Ki, map x/Kx, LYaw , i = 1 .. nops LYaw
odePsi d evalf subs
Iψ = 882500, β = convert 35,'units','deg','rad' , params, YForce, Iψ
$diff ψ t , t$2 = add x, x = zip `*`, LYaw, seq Fy
Ctan β $θ t $add x, x = map x/x$x, LYaw
8.82500 10
5
2
d
ψ t
dt2
= 11362.29826
d
dt
d
C1154.265220
dt
d
C1154.265220
dt
d
C2308.530440
dt
d
C11362.29826
dt
K9468.984961 θ t
C2308.530440
:
d
v1 t
dt
i t , i = 1 ..12
K11362.29826
d
v4 t
dt
d
v5 t K1154.265220
v6 t
dt
d
v7 t K1154.265220
v8 t
dt
d
v9 t K2308.530440
v10 t
dt
d
v11 t K11362.29826
v12 t
dt
v3 t
Kψ t
d
v2 t
dt
(32)
K2308.530440
K13523.1120 ψ t
The closed-loop system equations defining the behavior of the entire Maglev train in terms of heave, sway,
pitch and yaw is obtained by combining the twelve magnet module equations and the four equations of
motion defined above:
sysOverall d DynamicSystems NewSystem
op allcoils , odeY, odeZ, odeTheta, odePsi ,
inputvariable = Fd t , seq g i t , i = 1 ..12 , outputvariable = y t , z t , θ t , ψ t
;
Diff. Equation
continuous
(33)
inputvariable = Fd t , g1 t , g2 t , g3 t , g4 t , g5 t , g6 t , g7 t , g8 t , g9 t ,
g10 t , g11 t , g12 t
outputvariable = y t , z t , θ t , ψ t
3.2 Response Plots
The heave response plots to a vertical gust of wind and a descending guideway slope of 0.1ft/sec are
shown in Figures 8 and 9, respectively. From the plots we see that the control system developed for each
magnet is able to work together to minimize the heave displacement.
23 of 25
Heave Response
Response to Vertical Gust of Wind
8000
6000
4000
2000
0
0
2
4 6 8
time [secs]
10
Heave response to a gust of
wind
0.003
eave Displacement [in
Wind [lbs]
Input - Vertical Gust of Wind
10000
0.002
0.001
0
0
2
4
6
8
time [secs]
10
Figure 8B: Input Signal
Response to Descending Guideway
K6
K8
K10
K12
Heave response to descending
guideway
0
2 4 6 8 10 12
Time [secs]
K1. # 10 - 8
ve displacement [inch
Slope [inches]
Input - Descending Guideway
0
2
4
6
8 10
K2
time [secs]
K4
K2. # 10 - 8
K3. # 10 - 8
Figure 9B: Input Signal
24 of 25
Results
This worksheet documented the design and development of a controller to guide a EMS Maglev train along a
guideway. In section 1, we determined a mathematical model to describe the magnet dynamics of the lifting
magnets. In section 2, we developed a PID controller with feedback acceleration to maintain an air gap
between the train and the guideway of 2 inches and thus counteract the inherent unstable nature of the lifting
magnets (i.e. when there is no current in the coils the vehicle will drop under the influence of gravity causing
the undercarriage to hit the guideway and when the current in the magnet is very large, the magnet will be
attracted to the rail and hit the rail). We showed that the behavior of our controller adhered well to a set control
criteria by investigating the system response to disturbance forces and guideway changes. In section 3, we
used the magnet model and control system equations to derive the equations of motion for the entire Maglev
system. These equations were then used to examine the Maglev train's response to disturbance forces and
guideway changes. We were able to show that the control system defined in section 2 was able to control the
motion of the train in terms of heave displacement.
Legal Notice: The copyright for this application is owned by Maplesoft. Maplesoft is not responsible for any errors
contained within and is not liable for any damages resulting from the use of this material.
25 of 25

Optimizing the Controller Design to Guide the Motion of a Maglev Train

Transcription

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